Quantitative Methods - Quantitative Methods Section 2

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66. Vicky Walters has to conduct a series of tasks in order to complete her research project. The first task can be done in 6 different ways, the second one in 2 different ways, and the final task in 2 different ways. The total number of ways in which Walters can carry out all three tasks is most likely:

  • Option : B
  • Explanation : Use the multiplication rule of counting to determine the total number of ways the three tasks can be done. Total number of ways = 6 * 2 * 2 = 24.
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67. X is a discrete random variable with four possible outcomes: X= {1,2,3,4}. Which of the following best represent the probability function f(x) for the discrete variable X?

  • Option : B
  • Explanation : The sum of f(x) over all values of X must equal 1 and 0 <= p(x) <=1
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68. The probability function for a discrete random variable is denoted by g(y) = g(Y = y). Which of the following is most likely true?

  • Option : A
  • Explanation : The sum of the probabilities g(y) over all values of Y equals 1. The probability g(y) is a number between 0 and 1.
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69. A discrete uniform distribution consists of the following 12 values: 2.0, 6.2, -1.5, 2.4, 9.0, 4.1, -3.2, -1.0, 5.5, 8.2, 4.5 and 0.8 The probability of a value lying between -3.0 and 1.0 in a single draw from the distribution is closest to:

  • Option : B
  • Explanation : First order the values from smallest to largest. Then note that three of the twelve values are between -3.0 and 1.0. Thus, the probability of a draw from the distribution being between -3.0 and 1.0 is 3/12 = 0.25.
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70. A six sided biased dice has the probability of landing on its edge twice out of 50 throws. The probability of any number showing up is equal. When the dice is rolled, the prize is equal to the number it lands on, i.e. $1 for showing 1, $2 for showing 2 and so on. The prize of landing on its edge is $10. What is the expected value of the prize on a single roll of dice?

  • Option : A
  • Explanation : The probability of the dice landing on its edge is 2/50 = 0.04. Since the dice is biased, the probability of showing up any number won’t be 1/6. It has been stated that the probability of showing any number is equal, i.e. = 0.16. The distribution is as follows:
    Outcome ProbabilitiesPrizesExpected
    Value
    Edge   0.04$100.4
    1  0.16$1 0.16
    20.16 $20.32
    30.16$30.48
    4   0.16$40.64
    0.16$50.80
    6   0.16$60.96
       3.76
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